Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define pll pair<LL,LL>
#define x first
#define y second
#define pi acosl(-1)
#define sqr(x) ((x)*(x))
#define pdd pair<double,double>
#define MEMS(x) memset(x,-1,sizeof(x))
#define MEM(x) memset(x,0,sizeof(x))
#define less Less
#define EPS 1e-4
#define arg ARG
#define cpdd const pdd
#define rank Rank
#define MXN 200005
int main(){ 
    int p[200005];
    int n;
    scanf("%d",&n);
    if(n==1){
        printf("0\n");
        return 0;
    }
    for(int i=0;i<n;i++){
        int x;
        scanf("%d",&x);
        p[x]=i;
    }
    int ans=0;
    int last=0;
    if(p[n-1]>p[n])ans++,last=n;
    else last=n-1;
    for(int i=n-2;i>=1;i--){
        if(ans==0){
            if(p[i]>p[i+1])ans++,last=i+1;
        }
        else if(p[i]<p[i+1]&&(p[last]<p[i]||p[last]>p[i+1])){
          //  if(p[last]>p[i+1])
        //    last=i;
        }
        else if(p[i]>p[i+1]&&(p[i]>p[last]&&p[last]>p[i+1])){
            //last=i+1;
        }
        else{
            ans++;
            last=i+1;
        }
    }
    printf("%d\n",ans);
}
/**/